Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\cos \pi xdx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_0^\infty {\cos \pi xdx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\cos \pi xdx} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{\pi }\sin \pi x} \right]_0^b \cr
& = \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left[ {\sin bx - \sin 0} \right] \cr
& = \frac{1}{\pi }\mathop {\lim }\limits_{b \to \infty } \left[ {\sin bx} \right] \cr
& {\text{Evaluate the limit}} \cr
& = \frac{1}{\pi }\sin \left( \infty \right) \cr
& {\text{The }}\mathop {\lim }\limits_{x \to \infty } \sin x{\text{ does not exist, so the integral diverges}} \cr} $$
